半径 10x^2+10y^2-10x-y-137=0

解答

半径

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

r = \frac{5581}{20}

求解步骤

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

将

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

改写为圆标准方程

(x - \frac{1}{2})^{2} + (y - \frac{1}{20})^{2} = (\frac{5581}{20})^{2}

因此圆的性质是:

(a, b) = (\frac{1}{2}, \frac{1}{20}), r = \frac{5581}{20}

半径是:

r = \frac{5581}{20}

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